Question

i need help like RN!!!
what are the foci of the ellipse? graph the ellipse. 18x^2+36y^=648

Answer 1

Hi,

$18x^2+36y^2=648\\ \Rightarrow ( \dfrac{x}{6})^2+( \dfrac{y}{3* \sqrt{2} } )^2=1\\ a=6,\ b=3* \sqrt{2}\\ c= \sqrt{a^2-b^2} = \sqrt{36-18} =3* \sqrt{2}$
Center is (0,0)
Foci are (-3√2,0) and (3√2,0)

Answer 2

Answer:

The equation of ellipse is given by:

$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$                 ....[1]

where,

a and b are the semi major axis and semi minor axis.

Foci of ellipse = $(\pm c, 0)$

where, $c = \sqrt{a^2-b^2}$

As per the statement:

Given the ellipse is:

$18x^2+36y^2 = 648$

Divide both sides by 648 we have;

$\frac{x^2}{36}+\frac{y^2}{18} =1$

On comparing with [1] we have;

$a^2 =36$ and $b^2=18$

First find c:

$c = \sqrt{36-18}=\sqrt{18}=3\sqrt{2}$

Foci of the ellipse are:

$(\pm 3\sqrt{2}, 0)$

Therefore, the foci of the ellipse are, $(3\sqrt{2}, 0)$ and $(-3\sqrt{2} , 0)$